1. Moment

2. In addition to the tendency to move a body in the direction of its application, a force can also tend to rotate a body about an axis. The axis may be any line which neither intersects nor is parallel to the line of action of the force. M=Fd

3. Moment about a Point Point of application of a force and line of action of a force are as in Fig: F line of action of a force point of application The magnitude of the moment or tendency of the force to rotate the body about the axis perpendicular to the plane of the body is proportional both to the magnitude of the force and to the moment arm d, which is the perpendicular distance from the axis to the line of action of the force.

4. In some two-dimensional and many of the three-dimensional problems to follow, it is convenient to use a vector approach for moment calculations. Moment: Vector Definition

5. Clockwise and Counterclockwise moments: If tendency of the force is to rotate its moment arm in a counter clockwise manner it is called counterclockwise moment (CCW). We will assume CCW moment as a positive moment (+). If tendency of the force is to rotate its moment arm in a clockwise manner it is called clockwise moment (CW). We will assume CW moment as a negative moment (-).

6. Zero moments - If line of action of the force intersects the axis of rotation, there is no moment of the force about this axis of rotation - If line of action of the force is parallel to the axis of rotation, there is no moment of the force about this axis of rotation

7. Varignon’s Theorem The moment of a force about any point is equal to the sum of the moments of the components of the force about the same point.

8. Problems For the figure shown below, find moment of 400 N force about point A using: 1.Scalar approach 2.vector approach. x A O 3 m 45 0 y 400 N 6 m

9. x A O 3 m 45 0 y 400 N 6 m Scalar Approach

10. x A O 3 m 45 0 y 400 N 6 m (0,-6) (-3,0) Vector Approach

12. Alternative Solution (Using Varignon’s Theorem): Assuming that the counterclockwise moment is positive; M B = - F x.(0.16m) - F y.(0.2m) where F x = 400 N and F y = 693 N M B = -400N(0.16m)-693N(0.2m) = -202.6 N.m We can say M B = -203 N.m Since we assumed that the counterclockwise moment is positive and we obtained a negative moment value, this implies that the direction of M B is actually clockwise. So; the result is exactly the same as before, i.e.